Given direction ratios are $(2,-1,2)$ and $(x, 3,5)$ We know that the angle between the lines whose direction ratios are $\left(a_{1}, b_{1}, c_{1}\right)$ and $\left(a_{2}, b_{2}, c_{2}\right)$ is
$\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$\Rightarrow \cos \frac{\pi}{4}=\frac{2 x-3+10}{\sqrt{4+1+4} \sqrt{x^{2}+9+25}}=\frac{2 x+7}{\sqrt{9} \cdot \sqrt{x^{2}+34}}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{2 x+7}{3 \sqrt{x^{2}+34}} \Rightarrow 2 x+7=3 \sqrt{\frac{x^{2}+34}{2}}$
$\Rightarrow 4 x^{2}+49+28 x=\frac{9\left(x^{2}+34\right)}{2}$ (Squaring on both
sides)
$\Rightarrow 2\left(4 x^{2}+49+28 x\right)=9 x^{2}+306$
$\Rightarrow 8 x^{2}+98+56 x=9 x^{2}+306$
$\Rightarrow x^{2}-56 x+208=0$
$\therefore \mathrm{x}=\frac{56 \pm \sqrt{3136-812}}{2}=\frac{56 \pm 48}{2}=28 \pm 24$
$=4,52 .$
Smaller value $=4$.